Question

write ratios for sides.
2.
a
opp
12
c
5
b
hyp
13
sin a =
cos a =
tan a =
sin b =
cos b =
tan b =
3.
a
c
24
b
hyp
10
sin a =
cos a =
tan a =
sin b =
cos b =
tan b =
15

Explanation:

Step1: Recall trigonometric - ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: For triangle in problem 2

For $\angle A$:

• $\sin A=\frac{BC}{AB}=\frac{5}{13}$
• $\cos A=\frac{AC}{AB}=\frac{12}{13}$
• $\tan A=\frac{BC}{AC}=\frac{5}{12}$

For $\angle B$:

• $\sin B=\frac{AC}{AB}=\frac{12}{13}$
• $\cos B=\frac{BC}{AB}=\frac{5}{13}$
• $\tan B=\frac{AC}{BC}=\frac{12}{5}$

Step3: For triangle in problem 3

First, find the hypotenuse $AB$ using the Pythagorean theorem $AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{24^{2}+10^{2}}=\sqrt{576 + 100}=\sqrt{676}=26$.
For $\angle A$:

• $\sin A=\frac{BC}{AB}=\frac{10}{26}=\frac{5}{13}$
• $\cos A=\frac{AC}{AB}=\frac{24}{26}=\frac{12}{13}$
• $\tan A=\frac{BC}{AC}=\frac{10}{24}=\frac{5}{12}$

For $\angle B$:

• $\sin B=\frac{AC}{AB}=\frac{24}{26}=\frac{12}{13}$
• $\cos B=\frac{BC}{AB}=\frac{10}{26}=\frac{5}{13}$
• $\tan B=\frac{AC}{BC}=\frac{24}{10}=\frac{12}{5}$

Answer:

Problem 2:
$\sin A=\frac{5}{13}$, $\cos A=\frac{12}{13}$, $\tan A=\frac{5}{12}$, $\sin B=\frac{12}{13}$, $\cos B=\frac{5}{13}$, $\tan B=\frac{12}{5}$
Problem 3:
$\sin A=\frac{5}{13}$, $\cos A=\frac{12}{13}$, $\tan A=\frac{5}{12}$, $\sin B=\frac{12}{13}$, $\cos B=\frac{5}{13}$, $\tan B=\frac{12}{5}$